Is Partial Fraction Decomposition Needed for This Integral?

[JayKo]

[SOLVED] Integration of Inverse of f(x)

$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

$$\frac{1}{4y^{3}-18y^{2}+10y}$$(ln|$${y^{4}-6y^{3}+5y^{2}}$$|)is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks

 

[mjsd]

yes, factorise and then partial fractions is ok

 

[JayKo]

wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks

 

[mjsd]

firstly you can further factorise y^2-6y+5

 

[JayKo]

$$\frac{1}{y^{4}-6y^{3}+5y^{2}}$$

=$$\frac{1}{y^{2}(y^{2}-6y+5)}$$

=$$\frac{1}{y^{2}(y-1)(y-5)}$$

So, by Partial fraction, is this correct?

$$\frac{1}{y^{2}(y-1)(y-5)}$$
=$$\frac{A}{y}$$+$$\frac{B}{y^{2}}$$+$$\frac{C}{y-1}$$+$$\frac{D}{y-5}$$

and then, by comparing coefficient of y^3, y^2, y.

(A+C+D){y^{3}}=0
(-6A+B-5C-D){y^{2}}=0
(5A-6B)y=0
answer is A=6/25, B=1/5, C=3/10. D=-87/50.

but doubt my answer. anyone can point out?thanks

 

[JayKo]

firstly you can further factorise y^2-6y+5

i know, is just that i m slow using teX, please gimme a min.thanks

 

[mjsd]

I can tell you the answer...


no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly

 

[JayKo]

after i substitute A,B,C,and D.

i can get back my original 1/f(x)

 

[mjsd]

looks ok so far

 

[JayKo]

I can tell you the answer...


no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly

lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.

 

[mjsd]

FYI I've got
A=6/25
B=1/5
C=-1/4
D=1/100

 

[JayKo]

looks ok so far

if so, then
$$\frac{1}{y^{2}(y-1)(y-5)}$$=$$\frac{6}{25y}$$+$$\frac{1}{5y^{2}}$$+$$\frac{3}{10(y-1)}$$-$$\frac{87}{50(y-5}$$

 

[JayKo]

FYI I've got
A=6/25
B=1/5
C=-1/4
D=1/100

OOops,gimme a minute,i recheck :)

 

[JayKo]

FYI I've got
A=6/25
B=1/5
C=-1/4
D=1/100

you are right! :D :D :D thanks a lot.for pointing it out !:D

 

[JayKo]

$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right?

but this problem is actually make y the subject in this equation, how to go about it further?
$$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy = $$\int\$$ dtthe original question is $$\frac{dy}{dt}$$=$$y^{4}-6y^{3}+5y^{2}$$ find y(t).if $$\int\frac{1}{y^{4}-6y^{3}+5y^{2}}$$ dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y)
then i m stuck to make y the subject. anyone to point out the mistake.thanks

 

[JayKo]

anyone, can show me the right way?

 

[JayKo]

another differential eq. question.

y'=1+xy find y. how to go about it. i m clueless
my method of multiply dx to (1+xy) doesn't work in this case.

 

[rohanprabhu]

You can solve this by the solution for Linear Differential equation. The equation you have is:

$$\frac{dy}{dx} = 1 + xy$$

which can be written as:

$$\frac{dy}{dx} + y(-x) = 1$$

which is similar to:

$$\frac{dy}{dx} + yP = Q$$

Here, $P = (-x); Q = 1$

So, you have:

$$I.F = e^{\int (-x)dx}$$
$$I.F = e^{-\frac{x^2}{2}}$$

And hence, your solution is given by:

$$ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C$$

But, it's going to be a real pain with this:

$$\int e^{\frac{-x^2}{2}}dx$$

a method to which i can't think of right now. Maybe there's some other way..

 

[Gib Z]

That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.

 

[JayKo]

You can solve this by the solution for Linear Differential equation. The equation you have is:

$$\frac{dy}{dx} = 1 + xy$$

which can be written as:

$$\frac{dy}{dx} + y(-x) = 1$$

which is similar to:

$$\frac{dy}{dx} + yP = Q$$

Here, $P = (-x); Q = 1$

So, you have:

$$I.F = e^{\int (-x)dx}$$
$$I.F = e^{-\frac{x^2}{2}}$$

And hence, your solution is given by:

$$ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C$$

But, it's going to be a real pain with this:

$$\int e^{\frac{-x^2}{2}}dx$$

a method to which i can't think of right now. Maybe there's some other way..

great, this is something new to me, just started my linear algebra course.thanks.will study your solution

 

[mjsd]

sorry, I shouldn't have marked your post as "solved" so early yesterday.

 

[JayKo]

That integral evaluates to the error function. No "method" for that one, you just have to know it. Haven't checked the rest of your post though, so not sure if the rest of your solution is right.

thank for pointing it out. ;)

 

[JayKo]

sorry, I shouldn't have marked your post as "solved" so early yesterday.

is alright, i was thinking of tackle the question part by part. tomorrow is my tutorial lesson, will find out more with tutor. thanks

 

[rohanprabhu]

great, this is something new to me, just started my linear algebra course.thanks.will study your solution

np. however, as I've seen in your posts... try not to make consecutive posts in a discussion. i.e. try to put all the thing u want to say once in one post only. When someone replies to that post, then go ahead and make a new post in reply. This just keeps the forum clean. No offense.

 

[JayKo]

np. however, as I've seen in your posts... try not to make consecutive posts in a discussion. i.e. try to put all the thing u want to say once in one post only. When someone replies to that post, then go ahead and make a new post in reply. This just keeps the forum clean. No offense.

i see, thanks for the advise.will take note.thanks for your effort in solving it as well